Append elements in the value field of a dictionary using comprehensions

Question

I have a list of elements, lets say:

y = [1, 3, 1, 5, 1]

And I would like to create a dictionary where:

Keys: are the elements in y
Values: is a list of the elements that appear before the Key in y

I attempted the following comprehension.

a={elem:y[i] for i, elem in enumerate(y[1:])}

However, since the value field in the dictionary is not a list, it only keeps the previous element in the last occurrence of the key.

In other words, for this example I get the following:

{3: 1, 1: 5, 5: 3}

Is there a way to do so using comprehensions ?

Note: I forgot to add the desired result.

{3: [1], 1: [3,5], 5: [1]}

"Keys: are the elements in y": not possible: you have duplicate elements. — Jean-François Fabre, Aug 27 '18 at 15:44
maybe you mean: `a={elem:y[:i-1] for i, elem in enumerate(y[1:len(y)])}` but you still have the duplicate key issue — Jean-François Fabre, Aug 27 '18 at 15:45
You already found out that this is *at best* very unintuitive to do via comprehensions. Why not use a loop? — MisterMiyagi, Aug 27 '18 at 15:45
@Jean-FrançoisFabre I guess he wants a list of each value before every occurrence of the same key: ``{3: [1], 1: [3, 5], 5: [1]}``. Keys being duplicates is the entire point. — MisterMiyagi, Aug 27 '18 at 15:47

Jean-François Fabre · Answer 1 · 2018-08-27T16:35:00.020

Your keys are duplicated, so you cannot create a dictionary with them (you'll lose the first elements).

So comprehensions are difficult to use (and inefficient, as stated by other comprehension answers here) because of the accumulation effect that you need.

I suggest using collections.defaultdict(list) instead and a good old loop:

import collections

y = [1, 3, 1, 5, 1]

d = collections.defaultdict(list)

for i,x in enumerate(y[1:]):
    d[x].append(y[i])  # i is the index of the previous element in y

print(d)

result:

defaultdict(<class 'list'>, {1: [3, 5], 3: [1], 5: [1]})

thithien · Answer 2 · 2018-08-27T16:05:55.093

1

Just for the fun of it. Here's a comprehension.

a = {y[i]: [y[x-1] for x in range(len(y)) if y[x]==y[i]] for i in range(1, len(y))}

>> {3: [1], 1: [3,5], 5: [1]}

Just note that it's too long and inefficient to be allowed in any practical program.

Using the defaultdict as Jean-François Fabre suggested in his answer below should be the proper way.

edited Aug 27 '18 at 16:05

answered Aug 27 '18 at 15:48

thithien

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that answer is worth a +1 because you're showing the way _not_ to do it efficiently and you're aware of it – Jean-François Fabre Aug 27 '18 at 16:34

score 1 · Answer 3 · answered Aug 27 '18 at 16:03

Use enumerate and set operations.

{value: set(y[:i]) - {value} for i, value in enumerate(y)}
Out: {1: {3, 5}, 3: {1}, 5: {1, 3}}

It's a bit ugly and inefficient because in your example it works out a new answer each time it encounters 1, but it works out right because the final time it does this is the final time it encounters 1.

Append elements in the value field of a dictionary using comprehensions

3 Answers3